Work Energy Theorem and Potential Energy violation?


by easwar2641993
Tags: energy, potential, theorem, violation, work
easwar2641993
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#1
Nov11-12, 11:11 AM
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I came across a rather confusing topic about Work Energy Theorem and Potential Energy applied in lifting a particle.I will be glad if anyone clears it for me.

Consider a particle at a height =0.Potential Energy is considered as zero at height=0.Now it is lifted to a position where height is h so that Potential Energy of the particle is increased by a value mgh where m is the mass of it,g is the gravitational acceleration.Or in other words energy possessed by the particle is increased from zero to mgh.Now the problem begins.
Net work done on the system W(net)=Change in Kinetic energy.But here initial and final kinetic energy of the particle is considered to be zero since the particle is stationary at initial and final positions.This means there is no change in energy of the particle.Because energy is something required to do work.But since net work is zero ,then energy of the particle should be constant.And I just can't apply energy function for this case
Potential energy U(final)-U(initial)=-W=K(initial)-K(final)
But W=0.Then how can I say potential energy change is zero since it is obvious that it possesses mgh energy?
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Doc Al
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#2
Nov11-12, 11:36 AM
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The work-energy theorem says that if you consider the work done by all forces (including gravity) it will equal the change in the kinetic energy (not total energy). And that's just what happens.

If you consider the work done by all forces except gravity, then the work done will equal ΔK + ΔU. (The work done by gravity is already included in the potential energy term.)
DaTario
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#3
Nov11-12, 12:31 PM
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The work that is equal to mgh is the work of the hand which lift the particle. The work done by the gravitational force is -mgh. The net work is zero as it should be since no kinetic energy was gained.

Best wishes,

DaTario

mihirviveka
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#4
Jan25-13, 07:34 AM
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Work Energy Theorem and Potential Energy violation?


Maybe we should view the situation in this way?

since the object was moved a certain distance by a force work was done on it so its kinetic energy has increased but since it stays at that height the kinetic energy becomes potential energy?
ZapperZ
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#5
Jan25-13, 07:40 AM
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Quote Quote by mihirviveka View Post
Maybe we should view the situation in this way?

since the object was moved a certain distance by a force work was done on it so its kinetic energy has increased but since it stays at that height the kinetic energy becomes potential energy?
That doesn't make any sense, since the height in the problem state by the OP has clearly changed! Otherwise, there's no work done in the first place against gravity.

Zz.
mihirviveka
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#6
Jan26-13, 12:07 AM
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this thing has been bothering me for a long time

what I had in mind was this situation:

If we throw a rock from the ground to the top of a building then it's kinetic energy has clearly increased but it just stays on the roof of the building so the increase in kinetic energy has become increase in potential energy.

What we need is the mathematical representation of this idea, can anyone please do that?
jtbell
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#7
Jan26-13, 01:04 AM
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Suppose that the rock has zero potential energy at ground level and 100 J of potential energy when it's on the roof of the building.

You (standing on the ground) throw the rock upwards and do 100 J of work on it. Before you start to throw it, it has 0 J PE and 0 J KE, for a total of 0 J mechanical energy. When it leaves your hand, it has 0 J PE and 100 J KE, for a total of 100 J mechanical energy which came from the work you did on it. (I'm assuming the rock doesn't change height significantly while it's in your hand, so its PE doesn't change while you throw it.) When it reaches the rooftop and comes to rest again, it has 100 J PE and 0 J KE, which is still a total of 100 J of mechanical energy.
mihirviveka
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#8
Jan26-13, 04:26 AM
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Quote Quote by jtbell View Post
Suppose that the rock has zero potential energy at ground level and 100 J of potential energy when it's on the roof of the building.

You (standing on the ground) throw the rock upwards and do 100 J of work on it. Before you start to throw it, it has 0 J PE and 0 J KE, for a total of 0 J mechanical energy. When it leaves your hand, it has 0 J PE and 100 J KE, for a total of 100 J mechanical energy which came from the work you did on it. (I'm assuming the rock doesn't change height significantly while it's in your hand, so its PE doesn't change while you throw it.) When it reaches the rooftop and comes to rest again, it has 100 J PE and 0 J KE, which is still a total of 100 J of mechanical energy.
great! I think this settles the issue


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